Estimating Population Parameters with Confidence Intervals: Prob & Stats Fall '04, Week 12, Study notes of Statistics

Download Estimating Population Parameters with Confidence Intervals: Prob & Stats Fall '04, Week 12 and more Study notes Statistics in PDF only on Docsity! MATH-1530- 04/07/15/17 Probability & Statistics Fall 2004 / Week 12 Using Confidence Intervals to Estimate Unknown Population Parameters I. Estimation of an unknown population proportion (p) The best choice for an estimator of an unknown population proportion is the sample proportion p-hat. Example: The Associated Press (Dec. 16, 1991) reported that, in a sample of 507 adult U.S. citizens, only 142 correctly described the Bill of Rights as the first ten amendments to the U.S. Constitution. WHAT is the variable of interest? It is whether or not a citizen correctly describes the Bill of Rights, a categorical variable. It doesn’t make sense to calculate a mean in this case, but we can calculate a proportion. Remember that there are certain assumptions that should be checked before we carry out any inference concerning proportions. If we can meet these assumptions, then the amount of confidence with which we are constructing the intervals should be fairly accurate. What are the 3 conditions that need to be met (see p. 475 in BPS)? A. First, calculate the sample proportion p-hat, and use it to find a 95% confidence interval for the true proportion of all U.S. adults that could give a correct description of the Bill of Rights. The formula you need is n pp zp )ˆ1(ˆ ˆ    B. The margin of error calculated in part A is about 4%. What size sample would we need to produce a 95% C.I. with a (smaller) margin of error of m = 0.025? Assume that we can use the result of the “pilot study” described above to give us a hint of the value of the proportion of “successes” in the population; that is, plug in the p-hat calculated above for p* in the formula given below. (NOTE: In this case we have specific instructions about the value of p* to use, but in a situation where we do not have a clue about the real population proportion, we should use 0.5 for p*.) Here is the formula to use           pp m z n 1 2 II. Estimation of an unknown population mean (mu) The most natural and intuitive choice for the estimator of mu is the sample mean x-bar, but it is also nice that the mathematics ensures that x-bar is unbiased and has low variability as an estimator of mu (take my word for it and I’ll spare you the mathematics). However, before we start estimating we must examine the distribution of variable X in the population as well as the sampling distribution of x-bar. Example: One of the “everyday” uses of estimation is to assure quality control in manufacturing and industrial settings in which samples are collected as often as every day or every shift in order to check that a certain process is running according to specification. We are given n = 16 measurements of a critical dimension (in mm) on a sample (needs to be an SRS) of auto engine crankshafts. WHAT is the variable of interest? It is some dimension of a crankshaft which is measured in millimeters, a quantitative variable. The process mean is supposed to be mu = 224.000 mm but can drift away from this target during production. We have reason to believe that individual measurements in the population are distributed Normally, but the first thing we should do is some exploratory data analysis for the sample. Let’s look at a histogram and the basic statistics. (Minitab Output) Descriptive Statistics for Crankshaft Dimension Variable N Mean Median TrMean StDev SE Mean Minimum Maximum Q1 Q3 dimension (mm) 16 224.002 223.988 224.00 0.0618 0.0155 223.902 224.120 223.96 224.05 Notice that the distribution of measurements in the sample is roughly symmetric and there do not appear to be any outliers. There is nothing here that contradicts our assumption of normality. We should proceed with the estimation. Also note that the estimator is x-bar, but the specific estimate is the specific value of the sample mean, 224.002 mm. A. We wish to give a 90% confidence interval for the process mean at the time these crankshafts were produced. Suppose for now that we know the standard deviation for the population of all crankshafts produced is sigma = 0.060. The formula we need to carry out this estimation is         n zx  . B. However, we cannot assume that the population standard deviation will always be known. There is another estimation technique that uses the t-Student distribution (instead of the Normal) that can help us overcome this obstacle. With the t-Student model and the sample standard deviation s the C.I. formula looks like this         n s tx where t* is determined not only by the level of confidence, but also by n – 1 degrees of freedom (just take the sample size and subtract one to find the proper row in the t-table). Notice that this formula calls for the sample mean x-bar (which is given in the Minitab output). It also calls for the sample standard deviation s and the sample size n. Plug those values into the formula and recalculate the 90% confidence interval. The critical value of t can be found in the t-table in the column located above the desired confidence level and in the row for n - 1 degrees of freedom. C. Also calculate a 95% confidence interval and a 99% confidence interval for mu using the formula based on the t distribution. Are these intervals narrower or wider than the one previously calculated in part B? 223.90 223.93 223.96 223.99 224.02 224.05 224.08 224.11 0 1 2 3 4 5 6 7 C1 F re q u e n cy Sixteen Crankshaft Measurements